题目描述
Given an integer array
nums
with possible duplicates, randomly output the index of a giventarget
number. You can assume that the given target number must exist in the array.Implement the
Solution
class:
Solution(int[] nums)
Initializes the object with the arraynums
.int pick(int target)
Picks a random indexi
fromnums
wherenums[i] == target
. If there are multiple valid i’s, then each index should have an equal probability of returning.
例子
例子 1
Input:
["Solution", "pick", "pick", "pick"]
[[[1, 2, 3, 3, 3]], [3], [1], [3]]
Output:[null, 4, 0, 2]
Explanation:Solution solution = new Solution([1, 2, 3, 3, 3]);
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
solution.pick(1); // It should return 0. Since in the array only nums[0] is equal to 1.
solution.pick(3); // It should return either index 2, 3, or 4 randomly. Each index should have equal probability of returning.
Constraints
1 <= nums.length <= 2 * 10^4
-2^31 <= nums[i] <= 2^31 - 1
target is an integer from nums.
At most 10^4 calls will be made to pick.
解题思路
这道题思路比较简单,由于要返回对应目标的下标,所以我们需要用一个哈希表来对对应元素的下标进行存储,这样查询的时候只需要常数时间。由于数组中有重复元素,因此应该用向量来存储下标,在返回时随机返回一个即可。代码如下:
#include <unordered_map>
#include <vector>
class Solution {
public:
Solution(std::vector<int>& nums) {
for (int i = 0; i < nums.size(); ++i) {
hmap_[nums[i]].push_back(i);
}
}
int pick(int target) {
std::vector<int>& indices = hmap_[target];
return indices[rand() % indices.size()];
}
private:
std::unordered_map<int, std::vector<int>> hmap_;
};
- 时间复杂度:
- constructor:
O(n)
- pick:
O(1)
- constructor:
- 空间复杂度:
O(n)
GitHub 代码同步地址: 398.RandomPickIndex.cpp
其他题目: GitHub: Leetcode-C++-Solution 博客: Leetcode-Solutions